Question

Integrate cos2x.log(cosx + sinx/cosx - sinx)dx

Answer 1

you can see your answer

Answer 2

Answer:

The required integral is

$I = \frac {sin2x}{2} log \frac{cosx + sinx}{cosx - sinx}-\frac{1}{2}log(sec2x) + c$

Step-by-step explanation:

Given integral is

$I = ∫cos2x.log( \frac{cosx + sinx}{cosx - sinx} )$

We use by parts method to solve this integral

$∫udv = uv - ∫vdu$

Here,using ILATE rule logarithm is taken as u and trigonometric function is taken as dv.Therefore we get,

$u = log( \frac{cosx + sinx}{cosx - sinx} ) \: \\ dv = cos2x = > v = ∫cos2x dx\\ = \frac{sin2x}{2}$

Substituting these in the by parts relation,we get the following

$I = log( \frac{cosx + sinx}{cosx - sinx} ) \frac{sin2x}{2} - ∫ \frac{sin2x}{2} \times ( \frac{cosx - sinx}{cosx + sinx} ) \times ( \frac{(cosx - sinx)( - sinx + cosx) - (cosx + sinx)( - sinx - cosx)}{(cosx - sinx) {}^{2} } ) dx$

By simplifying the above expression we get the following,

$I = \frac{sin2x}{2} log \frac{cosx + sinx}{cosx - sinx} - ∫tan2x.dx \\ = \frac {sin2x}{2} log \frac{cosx + sinx}{cosx - sinx} - \frac{1}{2} log(sec2x) + c$

Therefore,the required integral is

$I = \frac {sin2x}{2} log \frac{cosx + sinx}{cosx - sinx}-\frac{1}{2}log(sec2x) + c$

Where C is some integration constant

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